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HardyWeinberg equilibrium

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Genotype frequencies. red flowers: 20 = homozygotes = AA. pink flowers: 20 = heterozygotes = Aa ... If we combined the alleles at random (per Mendel), then genotype ... – PowerPoint PPT presentation

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Title: HardyWeinberg equilibrium


1
Hardy-Weinberg equilibrium
2
Hardy-Weinberg equilibrium
Is this a true population or a mixture? Is the
population size dangerously low? Has migration
occurred recently? Is severe selection occurring?
3
Quantifying genetic variation Genotype
frequencies red flowers 20 homozygotes
AA pink flowers 20 heterozygotes Aa white
flowers 10 homozygotes aa
4
Quantifying genetic variation Genotype
frequencies red flowers 20 homozygotes
AA pink flowers 20 heterozygotes Aa white
flowers 10 homozygotes aa N alleles
Genotype frequency
5
Quantifying genetic variation Genotype
frequencies red flowers 20 homozygotes
AA pink flowers 20 heterozygotes Aa white
flowers 10 homozygotes aa N 50
alleles 100 Genotype frequency 202010
6
Quantifying genetic variation Genotype
frequencies red flowers 20 homozygotes
AA pink flowers 20 heterozygotes Aa white
flowers 10 homozygotes aa N 50
alleles 100 Genotype frequency
202010 Allelic frequencies A a
7
Quantifying genetic variation Genotype
frequencies red flowers 20 homozygotes
AA pink flowers 20 heterozygotes Aa white
flowers 10 homozygotes aa N 50
alleles 100 Genotype frequency
202010 Allelic frequencies A red (20
20) pink (20) 60 (or 0.6) a white (10
10) pink (20 ) 40 (or 0.4)
8
Quantifying genetic variation Genotype
frequencies red flowers 20 homozygotes
AA pink flowers 20 heterozygotes Aa white
flowers 10 homozygotes aa N 50
alleles 100 Genotype frequency
202010 Allelic frequencies A red (20
20) pink (20) 60 p a white (10 10)
pink (20 ) 40 q
9
Reduce these frequencies to proportions Genotype
frequencies AA 20 or 0.4 Aa 20
0.4 aa 10 0.2 Allelic frequencies A
p 0.6 a q 0.4
10
Check that proportions sum to 1 Genotype
frequencies AA 20 or 0.4 Aa 20
0.4 AA Aa aa 1 aa 10 0.2 Allelic
frequencies A p 0.6 p q 1
a q 0.4
11
Genotype frequencies AA 20 or 0.4
Aa 20 0.4 AA Aa aa 1 aa 10
0.2 Allelic frequencies A p 0.6
p q 1 a q 0.4 If we combined the
alleles at random (per Mendel), then
genotype frequencies would be predictable by
multiplicative rule AA Aa aa
12
Genotype frequencies AA 20 or 0.4
Aa 20 0.4 AA Aa aa 1 aa 10
0.2 Allelic frequencies A p 0.6
p q 1 a q 0.4 If we combined the
alleles at random (per Mendel), then
genotype frequencies would be predictable by
multiplicative rule AA p x p p2 Aa p x
q x 2 2pq aa q x q q2
13
Genotype frequencies AA 20 or 0.4
Aa 20 0.4 AA Aa aa 1 aa 10
0.2 Allelic frequencies A p 0.6
p q 1 a q 0.4 If we combined the
alleles at random (per Mendel), then
genotype frequencies would be predictable by
multiplicative rule AA p x p p2 0.36 Aa
p x q x 2 2pq 0.48 aa q x q q2
0.16
14
Genotype frequencies AA 20 or 0.4
Aa 20 0.4 AA Aa aa 1 aa 10
0.2 Allelic frequencies A p 0.6
p q 1 a q 0.4 If we combined the
alleles at random (per Mendel), then
genotype frequencies would be predictable by
multiplicative rule AA p x p p2 0.36 Aa
p x q x 2 2pq 0.48 aa q x q q2
0.16
check sum of the three genotypes
must equal 1
15
AA p x p p2 0.36 Aa p x q x 2 2pq
0.48 aa q x q q2 0.16 Thus,
frequency of genotypes can be expressed as p2
2pq q2 1
16
AA AA Aa p, q Aa aa aa
parental generation gamete
offspring (F1) genotypes
frequencies genotypes
reproduction
17
Hardy-Weinberg (single generation)
observed allele expected genotypes
frequencies genotypes
AA AA Aa p, q Aa aa aa
calculated deduced
parental generation gamete
offspring (F1) genotypes
frequencies genotypes
reproduction
18
Why would genotype frequencies ever NOT be the
same as predicted from the allele frequencies?
observed expected
AA AA Aa p, q Aa aa aa
calculated deduced
19
Hardy-Weinberg equilibrium Observed genotype
frequencies are the same as expected frequencies
if alleles combined at random
20
Hardy-Weinberg equilibrium Observed genotype
frequencies are the same as expected frequencies
if alleles combined at random Usually true if -
population is effectively infinite - no
selection is occurring - no mutation is
occurring - no immigration/emigration is
occurring
21
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22
How do we evaluate all this???? (how close
must the data be to the ratios?)
23
How do we evaluate all this???? (how close
must the data be to the ratios?)
genotype observed expected
AA 19 Aa 57 aa 24 total 100
24
How do we evaluate all this???? (how close
must the data be to the ratios?)
p (219 57)/200 0.475 q (57 224)/200
0.525
genotype observed expected
AA 19 Aa 57 aa 24 total 100
25
How do we evaluate all this???? (how close
must the data be to the ratios?)
p (219 57)/200 0.475 q (57 224)/200
0.525
genotype observed expected
AA 19 22.5 Aa 57 50 aa 24
27.5 total 100 100
26
How do we evaluate all this???? (how close
must the data be to the ratios?) Chi square
?2 (observed expected)2
expected
?
dev. from exp.
(obs-exp)2 genotype observed expected
(obs-exp) exp AA 19
22.5 Aa 57 50 aa 24 27.5 total 100
100
27
How do we evaluate all this???? (how close
must the data be to the ratios?) Chi square
?2 (observed expected)2
expected
?
dev. from exp.
(obs-exp)2 genotype observed expected
(obs-exp) exp AA 19
22.5 -3.5 0.54 Aa 57 50
7 0.98 aa 24 27.5 -2.5 0.45 total 100
100 1.97 ?2
degrees of freedom 2 ( N 1)
28
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29
P gt0.25 (observed data not significantly
different) from expected data)
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